Let $f \in C^k(0,1)$ and assume that the $k$th derivative is $\alpha$Hölder continuous. Assume that $f(x) = 0$ in a fixed interval $(a,b) \subset (0,1)$. Can we characterize (or at least find some examples of) nonconstant functions $f$ as above such that $$f^{(k)}_{C^{0,\alpha}(0,1)} \le \Vert f \Vert_{L^1(0,1)},$$ where $g_{C^{0,\alpha}} = \sup_{x,y \in (0,1)} \frac{g(x)g(y)}{xy^\alpha}$?

2$\begingroup$ Take any smooth $f$ and add a sufficiently large constant to it. Or am I missing something? $\endgroup$– Mateusz KwaśnickiJan 24 at 23:00

1$\begingroup$ @MateuszKwaśnicki Sorry, I forgot to add the assumption that $f(x) = 0$ ona given interval $(a,b) \subset (0,1)$ (so we cannot just add a constant to make its $L^1$ norm larger without also making the $\cdot_{C^{0,\alpha}}$ seminorm larger) $\endgroup$– HiroJan 24 at 23:07
The answer is no.
Indeed, let $t:=\alpha\in(0,1)$ and $c:=\f\_1:=\f\_{L^1(0,1)}\in(0,\infty)$. Suppose that $f\in C^k(0,1)$ and $f=0$ on $(a,b)$, with $0\le a<b\le1$. Suppose that the inequality in question holds.
Then for $x\in[b,1]$ we have $f^{(k)}(x)=f^{(k)}(x)f^{(k)}(b)\le c(xb)^t\le c$, $f^{(k1)}(x)\le c(xb)\le c$, $\ldots$, $f(x)\le c$, $\int_b^1f\le c(1b)$.
Similarly, $\int_0^af\le ca$. So, $$c=\f\_1=\int_0^af+\int_b^1f\le ca+c(1b)<c,$$ a contradiction. $\Box$